Problem: If $x \diamond y = xy+4x-y$ and $x \bigtriangleup y = (8-x)(y)$, find $(-4 \diamond -6) \bigtriangleup 0$.
Answer: First, find $-4 \diamond -6$ $ -4 \diamond -6 = (-4)(-6)+(4)(-4)-(-6)$ $ \hphantom{-4 \diamond -6} = 14$ Now, find $14 \bigtriangleup 0$ $ 14 \bigtriangleup 0 = (8-14)(0)$ $ \hphantom{14 \bigtriangleup 0} = 0$.